3 Measure and Integration
Then f n is a simple function and positive whenever f is positive (the
latter follows from the fact that in this case f −1 [ Bn,k ] = ∅ for all k, n).
Moreover, f n ( x ) → f ( x ). Indeed, if x ∈ Ω, then there exists an n0 ∈ N
such that | f ( x )| ≤ 2n0 . Then | f n ( x ) − f ( x )| ≤ 2−n for all n ≥ n0 .
If f ≥ 0 then f n ≤ f n+1 . Indeed, if f n ( x ) = k2−n then f ( x ) ∈ [k2−n , (k +
1)2−n ). But then either
f ( x ) ∈ [(2k)2−(n+1) , (2k + 1)2−(n+1) ),
in which case f n+1 ( x ) = (2k)2−(n+1) = f n ( x ), or
f ( x ) ∈ [(2k + 1)2−(n+1) , (2k + 2)2−(n+1) ),
in which case f n+1 ( x ) = (2k + 1)2−(n+1) > f n ( x ).
In the case where K = C, we find sequences of simple functions ( gn )
and (hn ) that converge to Re f and Im f , respectively. Then we set f n :=
gn + ihn for all n ∈ N.
For example, this is
natural for stopping
times of random
processes.
There exist metrics
on R resp. [0, ∞]
such that B (R) resp.
B ([0, ∞]) is the Borel
σ-algebra for this
metric.
We sometimes prefer to work with a slightly more general notion of
measurable functions. It has technical advantages to allow functions to
take the values ∞ or −∞.
Remark 3.54 (Extended real line). We put R := {−∞} ∪ R ∪ {∞} which
we endow with the σ-algebra B (R), defined as σ(B (R) ∪ {{−∞}, {∞}}).
It follows that a function f : (Ω, Σ) → R is measurable if and only if
f −1 [{∞}], f −1 [{−∞}] ∈ Σ and f −1 [ A] ∈ Σ for all A ∈ B (R).
Similarly, B ([0, ∞]) is defined as σ (B ([0, ∞)) ∪ {{∞}}). We also remark that Proposition 3.53 generalizes to this situation.
Exercise 3.55. Let (Ω, Σ) be a measurable space and f n : Ω → R be measurable for all n ∈ N. Then the functions lim infn→∞ f n , lim supn→∞ f n ,
infn∈N f n and supn∈N f n are measurable.
Hint: Observe that { x ∈ Ω : sup{ f n ( x ) : n ∈ N} > a} =
S
n ∈N { x
∈ Ω : f n ( x ) > a }.
3.7 The Lebesgue integral
Given
a measure space (Ω, Σ, µ), we now introduce the Lebesgue integral
R
f dµ for suitable complex-valued, measurable functions that we call
‘integrable’. We proceed in several steps and first define the integral for
(real-valued) measurable functions f taking values in [0, ∞].
Definition 3.56. Let (Ω, Σ, µ) be a measure space. If f : Ω → [0, ∞] is a
simple function, with standard representation f = ∑nk=1 ak 1 Ak , one defines the Lebesgue integral of f by
Z
n
f dµ :=
∑ ak µ( Ak )
k =1
72
3.7 The Lebesgue integral
where, by convention, 0 · ∞ := 0 and ∞ + ∞ = ∞.
The following lemma shows that the Lebesgue integral for nonnegative
simple functions is positively linear and monotone.
Lemma 3.57. If (Ω, Σ, µ) is
R a measure space,R f , g : Ω R→ [0, ∞] are simple
functions and
λ
>
0,
then
λ f + g dµ = λ f dµ + g dµ. Moreover, if
R
R
f ≤ g, then f dµ ≤ g dµ.
n
Proof. If ∑nk=1 ak 1 Ak is the standard representation
R of f , then
R ∑k=1 λak 1 Ak
is the standard representation of λ f . Now λ f dµ = λ f dµ follows
from the definition of the integral. Now let λ = 1 and ∑m
l =1 bl 1 Bl be the
w
standard representation of g. If ∑ j=1 c j 1Cj is the standard representation
S
of f + g, then Cj = ak +bl =c j Ak ∩ Bl . Thus
Z
f dµ +
Z
n
g dµ =
∑
k =1
w
=
m
ak µ( Ak ) + ∑ bl µ( Bl ) =
l =1
∑ c j µ(Cj ) =
Z
n
m
∑ ∑ (ak + bl )µ( Ak ∩ Bl )
k =1 l =1
f + g dµ,
j =1
where we have used the finite additivity of µ and the fact that the sets
Ak ∩ Bl are pairwise disjoint.
Now assume that f ≤ g. Consider the function g − f , with the convention that ∞ − ∞ = 0 and ∞ − c = ∞ for all c ∈ [0, ∞). Then g − f is a
nonnegative, simple function. Hence,
Z
g dµ =
Z
( g − f ) + f dµ =
Z
g − f dµ +
Z
f dµ ≥
Z
f dµ,
since the integral of a nonnegative, simple function is clearly nonnegative.
Lemma 3.58. Let (Ω, Σ, µ) be a measure
R space and f : Ω → [0, ∞] be simple.
Then ν : Σ → [0, ∞] defined by ν( A) = 1 A f dµ is a measure on (Ω, Σ).
Proof. As 1∅ f is the zero function, the map ν clearly satisfies property
(M1). It remains to very (M2). So let ( Ak ) be a sequence of disjoint sets in
S
Σ. Define Bn := nk=1 Ak . Then
ν( Bn ) =
Z
1Bn f dµ =
Z
n
∑
k =1
1 Ak f dµ =
n Z
∑
n
1 Ak f dµ =
∑ ν ( A k ).
k =1
k =1
Clearly, Bn ⊂ Bn+1 for all n ∈ N and A := n∈N Bn = k∈N Ak . Since f
is a simple function, take f = ∑m
j=1 a j 1Cj to be its standard representation.
S
S
73
The convention
0 · ∞ = 0 ensures
that the Lebesgue
integral of the
zero function on
R is zero. Note
that λ(R) = ∞.
3 Measure and Integration
Observe that
ν( A) =
Z
1 A f dµ =
Z
m
∑ a j 1Cj ∩ A dµ
j =1
m
=
∑ a j µ(Cj ∩ A)
j =1
m
∑ a j µ(Cj ∩ Bn )
n→∞
= lim
j =1
= lim
n→∞
= lim
m
Z
∑ a j 1Cj ∩Bn dµ
j =1
Z
n→∞
∞
1Bn f dµ = lim ν( Bn ) =
n→∞
∑ ν( Ak )
k =1
In the previous calculation we used the definition of the integral for simple functions, that it is positively linear, the continuity of the measure µ,
and in the last step the identity ν( Bn ) = ∑nk=1 ν( Ak ) that was established
above. This proves (M2). Hence ν is a measure.
We can now define the Lebesgue integral by approximation for an arbitrary measurable function f : Ω → [0, ∞].
Definition 3.59. Let (Ω, Σ, µ) be a measure space and f : Ω → [0, ∞] measurable. One defines
Z
nZ
o
:
f dµ = sup
g dµ 0 ≤ g ≤ f , g simple .
We say that f is (Lebesgue) integrable, if
R
f dµ < ∞.
Example 3.60. If δa is the Dirac measure on (Ω, Σ) for an a ∈ Ω, then for
all measurable f : Ω → [0, ∞] we have
Z
f dδa = f ( a)
and f is integrable if and only if f ( a) < ∞.
Proof. First, let f be a simple function. Then δa ( f −1 [{t}]) = R1 if a ∈
f −1 [{t}], i.e. f ( a) = t and δa [ f −1 [{t}]] = 0 else. Thus in this case f dδa =
f ( a). Now let f : Ω → [0, ∞] be a measurable function. If g is a simple
function with g ≤ f , then
Z
Ω
g dδa = g( a) ≤ f ( a).
Taking the supremum over such g, it follows that
74
R
f dµ ≤ f ( a).
3.7 The Lebesgue integral
For the reverse inequality, let g( x ) := f ( a)1 f −1 [{ f (a)}] ( x ). Since f is
measurable and { f ( a)} ∈ B (R), it follows Rthat f −1 [{Rf ( a)}] ∈ Σ. So g is
a simple function. Moreover, g ≤ f . Hence, f dδa ≥ g dδa = f ( a).
Exercise 3.61. Consider the measure space (N, P (N), ζ ). A measurable
function f : N → [0, ∞] is a sequence ( an )n∈N = ( f (n))n∈N in [0, ∞].
Show that f is integrable if and only if ( an ) ∈ `1 and in this case,
∞
Z
N
f dζ =
∑ an .
n =1
Theorem 3.62 (Monotone convergence theorem). Let (Ω, Σ, µ) be a measure space and f n : Ω → [0, ∞] be measurable functions for all n ∈ N that are
monotonically
increasing
in n. Then f ( x ) := supn∈N f n ( x ) is measurable and
R
R
f dµ = supn∈N f n dµ.
Proof. As a consequence of Exercise 3.55, f is measurable.
One deduces from f n ≤ f that
sup
n ∈N
Z
f n dµ ≤
Z
f dµ.
It remains to show the converse inequality. To this end, let 0 ≤ ϕ ≤ f be
simple and β > 1. It suffices to show that
Z
ϕ dµ ≤ β sup
n ∈N
Z
f n dµ.
The sufficiency follows from first taking the supremum over all simple
0 ≤ ϕ ≤ f and then taking the infimum over all β > 1. Define Bn :=
S
{ x ∈ Ω : β f n ( x ) ≥ ϕ( x )}. Then Bn ∈ Σ, Bn ⊂ Bn+1 andR n∈N Bn = Ω.
Moreover, we have β f n ≥ ϕ on Bn . Define ν( A) := 1 A ϕ dµ for all
A ∈ Σ. Then ν is a measure by Lemma 3.58. Hence
Z
ϕ dµ = ν(Ω) = lim ν( Bn )
n→∞
= lim
Z
ϕ1Bn dµ
n→∞
≤ sup
Z
n ∈N
= β sup
n ∈N
β f n dµ
Z
f n dµ.
This establishes the converse inequality. Hence supn∈N
R
f n dµ =
R
f dµ.
Remark 3.63. Let f : Ω → [0, ∞] be measurable. Let ( f n ) be a sequence
of nonnegative measurable functions such that f n ↑ f pointwise. Note
75
Instead of sup we
could write lim in
the statement of
Theorem 3.62. Of
course this requires
to consider improper
limits, thereby for
example setting f ( x )
to ∞ where ( f n ( x ))
is unbounded.
3 Measure and Integration
that such a sequence exists by Proposition 3.53. It now follows from Theorem 3.62 that
Z
Z
nZ
o
f dµ = sup
ϕ dµ : 0 ≤ ϕ ≤ f simple = sup f n dµ.
n ∈N
So while the integral for nonnegative measurable functions is defined as
the supremum over a very large class of simple functions, it is actually
already obtained by the supremum over an approximating sequence as
in Proposition 3.53.
Corollary 3.64. Let
space,
R (Ω, Σ, µ) be a measure
R
R f , g : Ω → [0, ∞] measurable
and
R λ > 0.R Then λ f + g dµ = λ f dµ + g dµ. Moreover, if f ≤ g, then
f dµ ≤ g dµ.
Proof. Let ( f n ) and ( gn ) be sequences of simple functions such that f n ↑ f
and gn ↑ g. Then λ f n + gn ↑ λ f + g. Using monotone convergence and
Lemma 3.57, we obtain
Z
λ f + g dµ = lim
Z
n→∞
= lim λ
n→∞
λ f n + gn dµ
Z
f n dµ +
Z
gn dµ = λ
Z
f dµ +
Z
g dµ
The second assertion is clear from the definition and was already observed in the proof of Theorem 3.62.
Note that the notion
of a null set crucially
depends on the
measure, of course.
So sometimes it is
necessary to point
out this dependence
by speaking about a
µ-null set.
Definition 3.65. Let (Ω, Σ, µ) be a measure space. A null set is a set
N ⊂ Ω such that there exists a set M ∈ Σ with N ⊂ M and µ( M ) = 0.
Now let P = P( x ) be a property which, depending on x, may be true or
false. We say that P holds almost everywhere or for almost every x ∈ Ω,
if { x ∈ Ω : P( x ) is false} is a null set. If µ is a probability measure, we
also say that P holds almost surely if it holds almost everywhere.
Remark 3.66. Note that we do not assume that a null set N is measurable.
A measure space in which all null sets are measurable is called complete.
It is not hard to see that the measure space (Ω, M , µ∗ ) in the proof of
Carathéodory’s theorem, Theorem 3.42, is always complete.
We note that there is a straightforward procedure to complete a measure space by minimally enlarging the σ-algebra and thus the domain of
the measure to make all null sets measurable.
Example 3.67. Consider (R, B (R), δ0 ). Then almost every (or more precisely, δ0 -a.e.) x ∈ R is equal to 0.
Consider (R, B (R), λ). Then almost every (or more precisely, λ-a.e.)
x ∈ R is not equal to 0. In fact, almost every x ∈ R is irrational.
Corollary 3.68.
R Let (Ω, Σ, µ) be a measure space and f : Ω → [0, ∞] be measurable. Then f dµ = 0 if and only if f ( x ) = 0 for almost every x ∈ Ω.
76
3.7 The Lebesgue integral
Proof. If f is a simple function, then the assertion
R is obvious. In the general case, let a measurable f : Ω → [0, ∞] with f dµ = 0 be given. Let
( f n ) be an increasing sequence of simple functions converging
to f . Such
R
a sequence exists by Proposition 3.53. It follows that f n dµ = 0 for
all n ∈ N. Hence, by the case above, { x ∈ Ω : f n ( x ) 6= 0} is a null set,
whence there exists Mn ∈ Σ with µ( Mn ) = 0 such that f n ( x ) = 0 for all
S
x 6∈ Mn . Put M := n∈N Mn . Then M ∈ Σ and µ( M) ≤ ∑∞
n=1 µ ( Mn ) = 0.
Moreover, x 6∈ M implies that f ( x ) = 0. Hence f = 0 almost everywhere.
If, conversely, f = 0 almost everywhere, then there exists a measurable
set M with f ( x ) = 0 for all x 6∈ M. If g is a simple function with 0 ≤
g ≤ f then g−1 [{ x }] ⊂
R M for all x > 0. By the definition of the integral
for
R simple functions, g dµ = 0 and hence, since g ≤ f was arbitrary,
f dµ = 0.
Exercise 3.69. Let (Ω, Σ, µ) be a measure space and f : Ω → [0, ∞] be
measurable. Show that ν : Σ → [0, ∞], defined by
ν( A) :=
Z
A
f dµ :=
Z
1 A f dµ,
defines a measure on (Ω, Σ).
Moreover, show that if µ( A) = 0, then ν( A) = 0. In other words, ν
is absolutely continuous with respect to µ, which is usually denoted by
writing ν µ.
Remark 3.70. In the setting of Example 3.69 the function f is called the
density of ν with respect to µ.
In probability theory a density for a distribution is commonly taken
with respect to the Lebesgue measure. According to the Radon–Nikodym
theorem a probability measure P on B (R) has a density with respect to
the Lebesgue measure if and only if it is absolutely continuous with respect to the Lebesgue measure. Equivalently, this is the case if and only if
its distribution function F ( x ) := P((−∞, x ]) is an absolutely continuous
function. Not every continuous real function is absolutely continuous,
but every Lipschitz function is.
Theorem 3.71 (Fatou’s Lemma). Let (Ω, Σ, µ) be a measure space, f n : Ω →
[0, ∞] be measurable and set f ( x ) := lim infn→∞ f n ( x ). Then f is measurable
and
Z
Z
f dµ ≤ lim inf f n dµ.
n→∞
Proof. Note that f ( x ) = supk≥1 infn≥k f n ( x ). Let gk ( x ) := infn≥k f n ( x ).
Then gk is measurable by Exercise 3.55.
R Clearly gkR↑ f . So it follows from
monotone convergence that supk∈N gk dµ = Ω f dµ. On the other
hand,
Z
Z
Z
f dµ = sup gk dµ ≤ lim inf f n dµ.
k ∈N
n→∞
77
3 Measure and Integration
R
R
Indeed, for every n ≥ k we have gk ≤ f nRand thus gk dµR ≤ f n dµ.
Since this is true for all n ≥ k, we have gk dµ ≤ infn≥k f n dµ. By
taking the supremum over k ∈ N on both sides, the above inequality
follows.
Definition 3.72. Let (Ω, Σ, µ) be a measure
R space, f : Ω → K be measurable. Then f is called integrable if | f | dµ < ∞. We write f ∈
L 1 (Ω, Σ, µ).
If K = R, we set
Z
f dµ :=
Z
+
f dµ −
Z
f − dµ.
Note that if | f | is integrable then f + and f − are both integrable nonnegative functions. If K = C, we set
Z
f dµ =
Z
Re f dµ + i
Z
Im f dµ.
Note that if | f | is integrable, then Re f and Im f are both integrable realvalued functions.
If f is an integrable (real or complex) measurable function and A ∈ Σ,
we define
Z
Z
f dµ :=
f 1 A dµ
A
Lemma 3.73. Let (Ω, Σ, µ) be a measure space.
R
R
(a) For all integrable f , we have | f dµ| ≤ | f | dµ.
R
(b) If Rf is integrable and λ ∈ K, then λ f is integrable and λ f dµ =
λ f dµ.
R
(c) If
f
and
g
are
integrable,
then
f
+
g
is
integrable
and
f + g dµ =
R
R
f dµ + g dµ.
Remark 3.74. Note that (b) and (c) can be expressed by saying
R that the
integrable functions form a vector space and the map f 7→
f dµ is a
linear map from the integrable functions to K.
Proof. Let us first consider the case where K = R.
(a) We have
Z
Z
Z
Z
Z
Z
+
−
+
−
f dµ + f dµ = | f | dµ,
f dµ = f dµ − f dµ ≤
where we have used Corollary 3.64 in the last step.
R
R
(b) First note that by Corollary 3.64 one has |λ f | dµ = |λ| | f | dµ <
∞ if f is integrable. This proves that λ f is integrable whenever f is.
78
3.7 The Lebesgue integral
Now, if λ > 0, then (λ f )+ = λ f + and (λ f )− = λ f − . Thus, using
Corollary 3.64,
Z
λ f dµ =
Z
=λ
Z
+
λ f dµ −
Z
+
f dµ − λ
λ f − dµ
Z
−
f dµ = λ
Z
f dµ.
If, on the other hand, λ < 0, then (λ f )+ = −λ f − and (λ f )− = −λ f + .
Thus, in this case,
Z
λ f dµ =
Z
=λ
−
−λ f dµ −
Z
+
f dµ − λ
Z
−λ f + dµ
Z
−
f dµ = λ
Z
f dµ.
(c) Since | f + g| ≤ | f | + | g|, it follows that
Z
| f + g| dµ ≤
Z
| f | + | g| dµ =
Z
| f | dµ +
Z
| g| dµ < ∞
if f and g are integrable.
Moreover, by definition and Corollary 3.64,
Z
f dµ +
Z
g dµ =
=
(∗)
=
=
Z
+
f dµ −
Z
Z
f
+
Z
−
f dµ +
+
+ g dµ −
+
+
g dµ −
Z
( f − + g− ) dµ
Z
( f + g)− dµ
( f + g) dµ −
Z
Z
Z
g− dµ
f + g dµ.
Here, (∗) follows from integrating the identity f + + g+ + ( f + g)− =
( f + g)+ + f − + g− and using Corollary 3.64.
R
In
the
case
where
K
=
C,
for
(b)
and
(c)
one
uses
that
f dµ =
R
R
Re f dµ + i Im f dµ. We omit the easy computations. For (a), we use
that for every complex number z one has |z| = supt∈R Re(eit z). So for
every t ∈ R one has
Re e
it
Z
f dµ = Re
Z
it
e f dµ =
Z
it
Re(e f ) dµ ≤
Z
| f | dµ.
Taking the supremum over t ∈ R, statement (a) follows.
Theorem 3.75 (Dominated convergence theorem). Let (Ω, Σ, µ) be a measure space and ( f n ) be a sequence of integrable functions with the following two
properties.
(a) f˜( x ) := limn→∞ f n ( x ) exists for almost every x ∈ Ω, say outside the set
N ∈ Σ with µ( N ) = 0.
79
3 Measure and Integration
The technicalities
with f and f˜ are
required as f˜ might
not be measurable.
Alternatively, to
avoid speaking
about f˜ and N,
one could assume
that a measurable
f : Ω → K is given
that is pointwise
almost everywhere
the limit of the f n .
(b) There exists an integrable function g with | f n ( x )| ≤ g( x ) for almost every
x ∈ Ω and all n ∈ N.
Then f : Ω → K, defined by f ( x ) = f˜( x ) if x 6∈ N and f ( x ) = 0 if x ∈ N, is
integrable and
lim
Z
n→∞
| f n − f | dµ = 0.
In particular,
lim
Z
n→∞
f n dµ =
Z
f dµ.
Proof. Changing f n and f on a set of measure zero, we may assume that
(a) and (b) hold everywhere. By Proposition 3.30, f is measurable. Since
| f | ≤ g, it follows that f is integrable.
Now observe that | f n − f | ≤ 2g and hence 2g − | f n − f | ≥ 0. By Fatou’s Lemma 3.71,
Z
2g dµ =
=
Z
lim inf(2g − | f n − f |) dµ ≤ lim inf
n→∞
Z
n→∞
2g dµ − lim sup
n→∞
Z
Z
2g − | f n − f | dµ
| f n − f | dµ.
R
R
So lim supn→∞ | f n − f | dµ = 0, and therefore limn→∞ | f n − f | dµ = 0.
By Lemma 3.73,
Z
Z
Z
f n dµ − f dµ ≤ | f n − f | dµ → 0.
This proves the claim.
Example 3.76. Let us give an example that condition (b) in Theorem 3.75
is necessary. Consider (R, B (R), λ). If we set f n := n1(0, 1 ) , then ( f n ) is
n
aR sequence of simpleR functions converging to 0 everywhere. However,
R f n dλ ≡ 1 6 → 0 = R 0 dλ.
In the lecture we
discussed here how
the expected value
of random variables
is usually computed
using densities for
the push-forward
measure.
Recall that µΦ ( A) =
µ(Φ−1 [ A]) for
all A ∈ F , see
Lemma 3.31.
Exercise 3.77. Consider the situation of Exercise 3.69, i.e.R (Ω, Σ, µ) is a
measure space, f : Ω → [0, ∞] is measurable and ν( A) := A f dµ.
Show that g is integrable with respect to ν if and only if g f is integrable
with respect to µ and in this case,
Z
g dν =
Z
g f dµ.
We close this section by considering the integration under a push-forward
measure, which is of great importance for applications.
Theorem 3.78. Let (Ω, Σ, µ) be a measure space, ( M, F ) be a measurable space
and Φ : (Ω, Σ) → ( M, F ) be measurable. We denote the push-forward of µ
under Φ by µΦ . Then for a measurable f : ( M, F ) → (K, B (K)) we have
80
3.8 On the connection between the Lebesgue and Riemann integral
f ◦ Φ ∈ L 1 (Ω, Σ, µ) if and only if f ∈ L 1 ( M, F , µΦ ). In this case,
Z
Ω
Z
f ◦ Φ dµ =
M
f dµΦ .
Proof. First, let f = ∑nj=1 ak 1 Ak be a nonnegative, simple function. Then
f ◦Φ =
n
∑ a k 1 Φ −1 [ A k ]
j =1
and thus, by the definition of the push-forward measure,
Z
Ω
f ◦ Φ dµ =
n
∑ ak µ(Φ
−1
n
[ A]) =
j =1
∑ ak µΦ ( A) =
Z
j =1
M
f dµΦ .
It follows that for nonnegative, simple functions the assertion holds true.
Now let f : M → [0, ∞] be measurable and ( f n ) be a sequence of simple
functions with f n ↑ f pointwise. Then, by monotone convergence and the
above,
Z
Ω
f ◦ Φ dµ = sup
Z
n ∈N Ω
f n ◦ Φ dµ = sup
Z
n ∈N M
f n dµΦ =
Z
M
f dµΦ .
This shows that the assertion holds for arbitrary measurable positive f .
Since | f ◦ Φ| = | f | ◦ Φ, it follows that f ∈ L 1 ( M, F , µΦ ) if any only
if f ◦ Φ ∈ L 1 (Ω, Σ, µ). The general formula follows by splitting real
valued functions f into the positive functions f + and f − and complex
valued functions f into Re f and Im f .
3.8 On the connection between the Lebesgue
and Riemann integral
We next compare the Lebesgue integral with the Riemann integral. As
is well-known, every continuous function f : [ a, b] → R is Riemann integrable. We now show that such functions are also Lebesgue integrable
and the Lebesgue integral agrees with the Riemann integral.
Theorem 3.79. If f : [ a, b] → K is continuous, then f is a Lebesgue integrable
function on ([ a, b], B ([ a, b]), λ). Moreover,
Z b
Z
[ a,b]
f dλ = R-
a
f (t) dt.
Proof. Let a sequence of partitions πn := (t0(n) , . . . , t(knn ) ) with |πn | → 0
be given and let ξ n = (ξ 1(n) , . . . , ξ k(nn ) ) be a sequence of associated sample
81
3 Measure and Integration
points. Put
kn
f n :=
∑ f (ξ (j n) )1[t(j−n)1,t(j n) ) .
j =1
R
Then f n is a simple function and [a,b] f n dλ = S( f , πn , ξ n ).
Moreover, (a) | f n | ≤ k f k∞ and the latter is integrable on our measure
space and (b) f n (t) → f (t) for all t ∈ [ a, b]. Indeed, for fixed t ∈ [ a, b], we
have | f n (t) − f (t)| = | f (ξ (jnn) ) − f (t)|, where ξ (jnn) is the sample point in
the interval [t(jnn−) 1 , t(jnn) ] and jn is chosen such that t lies in this interval. But
then |ξ (jnn) − t| ≤ |t(jnn) − t(jnn−) 1 | ≤ |πn | → 0 and hence, by the continuity of
f , it follows that | f (ξ (jnn) ) − f (t)| → 0.
Hence the dominated convergence theorem, Theorem 3.75, applies and
shows that f is integrable and
Z
[ a,b]
f dλ = lim
Z
n→∞ [ a,b]
f n dλ = lim S( f , πn , ξ n ).
n→∞
Rb
Since, on the other hand, the Riemann sums converge to R- a f (t) dt, the
assertion follows.
Remark 3.80. Actually, the continuity assumption in Theorem 3.79 is not
needed, if one is willing to enlarge the σ-algebra. It can be proved that if
f : [ a, b] → K is Riemann integrable then it is almost everywhere equal to
a measurable function that is Lebesgue integrable and the Riemann and
the Lebesgue integral coincide.
There is also an extension of Theorem 3.79 to improper Riemann integrals. We recall that if −∞ < a < b ≤ ∞ and f : [ a, b) → R is continuous, then
R rf is called improperly Riemann integrable on [ a, b) if the limit
limr↑b R- a f (t) dt exists. The limit is then called the improper Riemann
R b−
integral of f over [ a, b) and denoted by R- a f (t) dt.
Theorem 3.81. Let −∞ < a < b ≤ ∞ and f : [ a, b) → R be continuous,
R b−
such that the improper Riemann integral R- a | f (t)| dt exists, then f is inR b−
tegrable on ([ a, b), B ([ a, b)), λ), the improper Riemann integral R- a f (t)dt
exists and
Z
Z
b−
[ a,b)
f dλ = R-
a
f (t) dt.
Proof. Pick a sequence (bn ) ⊂ ( a, b) with bn ↑ b. By monotone convergence and Theorem 3.79,
Z
[ a,b)
| f | dλ = lim
Z
= lim Rn→∞
82
| f | dλ
n→∞ [ a,bn ]
Z bn
a
Z b−
| f (t)| dt = R-
a
| f (t)| dt < ∞.
3.8 On the connection between the Lebesgue and Riemann integral
It follows that | f | is integrable on [ a, b). Moreover, since 1[a,bn ) f converges
to f pointwise and |1[a,bn ) f | ≤ | f |, the dominated convergence theorem
yields
Z
[ a,b)
f dλ = lim
Z bn
Z
n→∞ [ a,bn )
f dλ = lim Rn→∞
a
Z b−
f (t) dt = R-
a
f (t) dt,
where we have used Theorem 3.79 in the second step.
Remark 3.82. Similar results as in Theorem 3.81 also hold for improper
Riemann integrals that are improper on the left-hand side or on both
sides.
R
Example 3.83. In Theorem 3.81, the assumption that R- [a,b) | f (t)| dt exists
is crucial and cannot be omitted. An example is given by f : [1, ∞) → R ,
defined by f (t) = sint t In this case, by integration by parts, we obtain
Z x
sin t
R-
1
t
dt =
Z x
Z ∞
− cos t x
cos t
cos t
−
Rdt
→
cos
1
−
Rdt
t
t2
t2
1
1
1
as x → ∞. The latter improper Riemann integral exists since |t−2 cos t| ≤
t−2 and the
is integrable. It follows that the improper Riemann
R ∞latter
cos t
integral R- 1 t2 dt exists.
On the other hand, on each interval [kπ, (k + 1)π ), we have | f (t)| ≥
|sin(t)|((k + 1)π )−1 . It thus follows that
n
Z
[1,∞)
1
∑ (k + 1)π [kπ,(k+1)π) |sin(t)| dλ(t)
k =1
Z π
1 n
1 =
· R- |sin(t)| dt.
π k∑
k+1
0
=1
Z
f dλ ≥
Since the harmonic series diverges, it follows that f is not integrable on
[1, ∞).
Remark 3.84. In what follows, we will also use the ‘differential’ dt in
Lebesgue integrals instead of the (formally correct) dλ. We will thus
write
Z
Z
b
[ a,b]
f (t) dt
or
a
f (t) dt
to denote the Lebesgue integral of f on the interval [ a, b]. This is particularly helpful when the function f depends on more than one variable.
To have
R this feature also at hand
R for general measures, we will frequently
write Ω f ( x ) dµ( x ) instead of Ω f dµ to emphasize that we are integrating with respect to the variable x.
83
3 Measure and Integration
3.9 Integrals depending on a parameter
Instead of [0, 1] we
could use a general
metric space here.
The main topic of this section is to interchange operations like integration, differentiation and taking limits. This is a topic at the very heart of
analysis.
Suppose that (Ω, Σ, µ) is a measure space. If we are given a map
f : [0, 1] × Ω →RC such that f (t, ·) is integrable for all t ∈ [0, 1], we may
define F (t) := Ω f (t, x ) dµ( x ). It is then natural to ask how F depends
on the parameter t ∈ [0, 1]. In this short section, we use the dominated
convergence theorem to prove some results in this direction.
Proposition 3.85. Let (Ω, Σ, µ) be a measure space. Furthermore, let f : [0, 1] ×
Ω → K be such that the following three properties hold.
(a) x 7→ f (t, x ) ∈ L 1 (Ω, Σ, µ) for all t ∈ [0, 1].
(b) t 7→ f (t, x ) is continuous for almost all x ∈ Ω.
(c) There exists a g ∈ L 1 (Ω, Σ, µ) such that | f (t, x )| ≤ g( x ) for all (t, x ) ∈
[0, 1] × Ω.
R
Then F : [0, 1] → C defined by F (t) = Ω f (t, x ) dµ( x ) is continuous.
Proof. Let tn → t in [0, 1]. Then f (tn , x ) → f (t, x ) for almost all x ∈ Ω by
(b). Since | f (tn , x )| ≤ g( x ) for all x ∈ Ω by assumption and g ∈ L 1 (Ω),
it follows from the dominated convergence theorem, Theorem 3.75, that
F (tn ) =
Z
Ω
f (tn , x ) dµ( x ) →
Z
Ω
f (t, x ) dµ( x ) = F (t).
This proves the continuity of F.
Proposition 3.86. Let I be an interval in R and (Ω, Σ, µ) be a measure space.
Furthermore, let f : I × Ω → K be such that the following three properties hold.
(a) x 7→ f (t, x ) ∈ L 1 (Ω, Σ, µ) for all t ∈ I.
(b) t 7→ f (t, x ) is differentiable for all x ∈ Ω.
(c) There exists a g ∈ L 1 (Ω, Σ, µ) such that | ∂t∂ f (t, x )| ≤ g( x ) for all
(t, x ) ∈ I × Ω.
R
Then F : I → K defined by F (t) = Ω f (t, x ) dµ( x ) is differentiable. Moreover,
∂
∂t f ( t, x ) is integrable for all t ∈ I and
F 0 (t) =
d
dt
Z
Ω
f (t, x ) dµ( x ) =
Z
Ω
∂
f (t, x ) dµ( x ).
∂t
Proof. Fix t ∈ I and let (tn ) be a sequence in I that converges to t. Define hn , h : Ω → C by hn ( x ) := (tn − t)−1 ( f (tn , x ) − f (t, x )) and h( x ) =
∂
∂t f ( t, x ). Then hn is integrable for every n ∈ N as a linear combination of
integrable functions. Moreover, hn ( x ) → h( x ) for all x ∈ Ω by assumption. By the mean-value theorem, hn ( x ) = ∂t∂ f (ξ n , x ) for some ξ n between
84
3.10 Product measures
t and tn . In particular, |hn | ≤ g. Thus the dominated convergence theorem shows that h is integrable and
F (tn ) − F (t)
=
tn − t
Z
Ω
hn ( x ) dµ( x ) →
Z
Ω
h( x ) dµ( x ) =
Z
Ω
∂
f (t, x ) dµ( x ).
∂t
This finishes the proof.
3.10 Product measures
In this section we construct a σ-algebra and a corresponding measure on
the product of two suitable measure spaces. Our motivation is to extend
the theory in order to deal with iterated integrals. The product measure
will allow us to write an iterated integral as a single integral with resprect
to the product measure.
Definition 3.87. Let (Ωk , Σk ) be a measurable space for k = 1, . . . , n.
The product (measurable space) of the spaces (Ωk , Σk ) is the measurable
N
space (∏nk=1 Ωk , nk=1 Σk ), where ∏nk=1 Ωi is the Cartesian product of the
N
sets Ωk , i.e., the set of all tuples ( x1 , . . . , xn ) where xk ∈ Ωk and nk=1 Σk
is generated by the cuboids A1 × · · · × An where Ak ∈ Σk .
Exercise 3.88. Let (Ω, Σ) and, for k = 1, . . . , n, also (Ωk , Σk ) be measure
spaces. Let f k : Ω → Ωk be a function and define f : Ω → ∏nk=1 Ωk by
N
f ( x ) = ( f 1 ( x ), . . . , f n ( x )). Show that f is Σ/ nk=1 Σk -measurable if and
only if f k is Σ/Σk -measurable for all k = 1, . . . , n.
In the following, let (Ωi , Σi , µi ) be σ-finite measure spaces for i = 1, 2.
We define a measure µ1 ⊗ µ2 on the σ-algebra Σ1 ⊗ Σ2 which is the product of the measures µ1 and µ2 in the sense that
µ1 ⊗ µ2 ( A × B ) = µ1 ( A ) µ2 ( B )
for all A ∈ Σ1 and B ∈ Σ2 . Note that as µ1 and µ2 are σ-finite, by Corollary 3.38 there exists at most one such measure.
For a set Q ⊂ Ω1 × Ω2 and x ∈ Ω1 , y ∈ Ω2 , we define the cuts [ Q] x
and [ Q]y by
[ Q] x := {y ∈ Ω2 : ( x, y) ∈ Q} and [ Q]y := { x ∈ Ω1 : ( x, y) ∈ Q}.
Lemma 3.89. For x ∈ Ω1 , y ∈ Ω2 and Q ∈ Σ1 ⊗ Σ2 we have [ Q] x ∈ Σ2 and
[ Q ] y ∈ Σ1 .
Proof. We put G := { Q ∈ Σ1 ⊗ Σ2 : [ Q] x ∈ Σ2 }. We claim that G is a σalgebra on Ω1 × Ω2 . Clearly (S1) holds, since [Ω1 × Ω2 ] x = Ω2 . (S2) and
85
Observe that the
2-dimensional
Lebesgue measure
λ2 is the product
measure of the
one-dimensional
Lebesgue measure with itself.